similar to: adding 1 month to a date

Hi, I have an array of 2d node locations and an array triangles, and would like to plot something like a image or persp. An example of doing it with rgl is: library(ncdf) library(rgl) # wget http://www.maplepark.com/~drf5n/extras/teapot.nc teapot<-open.ncdf("teapot.nc") z<-get.var.ncdf(teapot,"tris") l<-get.var.ncdf(teapot,"locations")

Hi, I'd like to partition a 2d probability density function into regions of equal probability. It is straightforward in the 1d case, like qnorm(seq(0,1,length=5)) but for 2d I'd need more constraints. Any suggestions for how to approach this? Is seems like a spatial sampling problem but I'm not sure where to look. Thanks for your time, Dave -- Dr. David Forrest drf at

Hello all, It seems that the next improvement to the R Graph Gallery is categorization of the graphics, that way each graph will be easier to find. That step should be done *carefully* if we want to avoid the opposite side-effect : graph not reachable through the categories. That's why the wisdom of the R community is required. Graphics will be classified in : - categories -

The following code produces an eps file with the tops of each of the ylabs clipped off. par(mfrow=c(2,2)) plot(runif(10), ylab="Function(Lengthy Expression)",xlab="Prediction") plot(runif(10), ylab=expression(Delta * Beta^2),xlab="Prediction") plot(runif(10), ylab="Function(Lengthy Expression)",xlab="Prediction")

My friend Veslot proposed me the Raven ! http://www.teteamodeler.com/allopass/images/corbeau.jpg As a matter of fact the intelligence of this bird is comparable to one of a monkey, although its brain is close to a reptile brain. R is quite the same : small, compact, but so clever. This is a canadian website explaining (in french) the R-aven intelligence

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I have a problem when using nnet to predict the negative values. For example : X = matrix(c(1,1,0,0,1,0,1,0),4,2) X Y = matrix(c(0,1,1,0)) - 0.5 # XOR - 0.5 Y nn = nnet(X,Y,size=3) val = predict(nn,X) val # this is expected to be close to Y, but it's not ! The 'val' is always positive. I tried to change the options, but the result isn't much better. Could someone give me an

I wish to perform brain surgery this afternoon at 4pm and don't know where to start. My background is the history of great statistician sports legends but I am willing to learn. I know there are courses and numerous books on brain surgery but I don't have the time for those. Please direct me to the appropriate HowTos, and be on standby for solving any problem I may encounter while

Hi, I failed to search for R package providing random number generator of "Park and Miller". Anyone know any R package supporting this kind of function? Thanks, Grace -- View this message in context: http://www.nabble.com/Random-Number-Generator-of-Park-and-Miller-tf3636501.html#a10154554 Sent from the R help mailing list archive at Nabble.com.

even for people who didn't like the outcome of the US elections this year, it must have been a joy to see all the nice (and not so nice) graphs that were shown. As an exercise, I recreated the map shown on the NY-Times website [ http://www.nytimes.com/packages/khtml/2004/11/03/politics/ 20041103_px_ELECT_GRAPHIC.html ] If you're interested,

Hi, According to help(sub), the ^ should match the zero-length string at the beginning of a string: sub('^','var',1:3) # "1" "2" "3" sub('$','var',1:3) # "1var" "2var" "3var" # This generates what I expected from the first case: sub('^.','var',11:13) # "var1" "var2"

Is there an element-by-element multiplication in R, like the .* operator in Matlab? eg: A (2x3) B (2x3) C=A.*B C (2x3) C = [[a11*b11 a12*b12 a13*b13]; [a21*b21 a22*b22 a23*b23]] I can't find one... Thanks -Mike Gates

Dear Sir/Madam, I am doing a project related to R. However, it is always difficult find some R functions. The R user guide seems not complete. Is there any free document about all R functions ? Who knows ? please help me. My email is myzhaogong@yahoo.com Thank you very much --------------------------------- [[alternative HTML version deleted]]

Hi All, I am new to post question on this list. I apologize if this question is too easy or irrelevant. I am doing a simulation study and I need to read a data file that can be easily read by SAS. So, what I try to do is to execute SAS in R and then read the output of SAS to R. I try the following code but it didn't work. system("c:\\program files\\sas institute\\v8\\sas.exe

Hello, I don't know if the result of acos(0.5) == pi/3 is a bug or not. It looks strange to me. Inaki Murillo

Hi, I need to be able to add one month or in some cases on eyear to a Date. Does anyone know of any easy ways of doing this? thanks scott -- Posted via http://www.ruby-forum.com/.

Hi: I am doing some work that involves plotting points of interest superimposed on photographs and maps. I can produce the plots fine in R, but so far I have had to do the superimposition externally, which makes it tedious to do exploratory work. I have looked to see if there is some capability to put a background picture on a plot window, but I have not found anything. Advice, anyone? Cheers

>>>>> "David" == David Forrest <drf5n@maplepark.com> >>>>> on Tue, 22 Mar 2005 15:02:20 -0600 (CST) writes: David> According to help(sub), the ^ should match the David> zero-length string at the beginning of a string: yes, indeed. David> sub('^','var',1:3) # "1" "2" "3"

Good day, I read some data from a PostgreSQL database by a following script: library(Rdbi) library(RdbiPgSQL) # conn becomes an object which contains the DB connection: conn <- dbConnect(PgSQL(), host="localhost", dbname="BVS", user="postgres", password = "*******") query_duj_kal <- dbSendQuery(conn, "select zdroj as well, cas as date, fe2,

Hi, I have a set of irregular time series and i want to produce a simple plot, with dates on x axis and attribute value of y axis. This is simple enough but my x axis is divided automatically by ticks every 5 years. I would like to have a tick every year at January 1st. I am not sure how i can do that - i end up with something very close to what i want, but it is clunky and not very correct. I